Free Math Worksheets

- List the favorable outcomes for each event before adding probabilities. - Remember that \(1\) is not prime. - Identify the perfect squares from \(1\) through \(8\). - For an “or” event, include every outcome satisfying at least one condition, without double-counting overlaps.

1. Event \(A\) contains \(1\) and \(4\), so \(P(A)=0.05+0.20=0.25\). 2. Event \(B\) contains \(5\), \(6\), \(7\), and \(8\), so \(P(B)=0.08+0.12+0.15+0.15=0.50\). 3. Event \(C\) contains the primes \(2\), \(3\), \(5\), and \(7\), so \(P(C)=0.15+0.10+0.08+0.15=0.48\). 4. The odd outcomes are \(1\), \(3\), \(5\), and \(7\), and the outcomes divisible by \(4\) are \(4\) and \(8\). These sets are disjoint, so \(P(D)=0.05+0.10+0.08+0.15+0.20+0.15=0.73\).

\(A\): \(P(A)=0.25\) \(B\): \(P(B)=0.50\) \(C\): \(P(C)=0.48\) \(D\): \(P(D)=0.73\)

- Recall the possible results of one die roll. - Translate “absolute difference” into an equation involving the two coordinates. - Check each possible first-roll value and determine which second-roll values make the difference \(2\). - Remember that order matters: \((1, 3)\) and \((3, 1)\) are different outcomes.

1. An outcome is an ordered pair \((a, b)\), where \(a, b \in \{1, 2, 3, 4, 5, 6\}\). 2. The condition \(X=2\) means \(|a-b|=2\). 3. Checking the possible first-roll values gives the pairs \((1, 3)\), \((2, 4)\), \((3, 1)\), \((3, 5)\), \((4, 2)\), \((4, 6)\), \((5, 3)\), and \((6, 4)\). 4. Therefore, \(E=\{(1, 3), (2, 4), (3, 1), (3, 5), (4, 2), (4, 6), (5, 3), (6, 4)\}\).

\(E=\{(1, 3), (2, 4), (3, 1), (3, 5), (4, 2), (4, 6), (5, 3), (6, 4)\}\)

- How many outcomes are possible when a coin is tossed three times? - Determine how many heads and tails make the difference “heads minus tails” equal to \(-1\). - Decide whether more than one arrangement has those numbers of heads and tails. - List every possible order.

1. There are \(2^3=8\) possible ordered triples of \(H\) and \(T\). 2. Let \(n_H\) be the number of heads and \(n_T\) the number of tails. Then \(n_H+n_T=3\) and \(n_H-n_T=-1\). 3. Substituting \(n_T=3-n_H\) gives \(n_H-(3-n_H)=-1\), so \(2n_H-3=-1\) and \(n_H=1\). Thus, the event contains outcomes with exactly one head and two tails. 4. The possible orders are \((H, T, T)\), \((T, H, T)\), and \((T, T, H)\). 5. Therefore, \(E=\{(H, T, T), (T, H, T), (T, T, H)\}\).

\(E=\{(H, T, T), (T, H, T), (T, T, H)\}\)

- What must the probabilities of all outcomes add to? - Write the relationship between the two unknown probabilities using one variable. - To find the probability of an event, add the probabilities of its outcomes.

1. Let \(P(x_4)=p\). Then \(P(x_3)=3p\). 2. Use the fact that all outcome probabilities add to \(1\): \(0.12+0.28+3p+p=1\). 3. Solve: \(0.40+4p=1\), so \(4p=0.60\) and \(p=0.15\). Therefore, \(P(x_4)=0.15\) and \(P(x_3)=0.45\). 4. Add the probabilities of the outcomes in \(E\): \(P(E)=0.12+0.45=0.57\).

\(P(x_3)=0.45\), \(P(x_4)=0.15\), and \(P(E)=0.57\).

- Represent the ordered outcomes in a grid. - For part b), subtract the overlap when adding the two events. - Identify the powers of \(2\) available on each die.

1. There are \(8\cdot12=96\) equally likely ordered outcomes. 2. A sum of \(10\) occurs for \((1, 9),(2, 8),(3, 7),(4, 6),(5, 5),(6, 4),(7, 3),(8, 2)\), so \(P=\frac{8}{96}=\frac{1}{12}\). 3. For at least one \(7\), use the addition rule: \(P=\frac{1}{8}+\frac{1}{12}-\frac{1}{96}=\frac{19}{96}\). 4. A product is a power of \(2\) only when each factor is a power of \(2\). Each die can show \(1,2,4,8\), giving \(4\cdot4=16\) favorable outcomes. Thus, \(P=\frac{16}{96}=\frac{1}{6}\).

a) \(\frac{1}{12}\) b) \(\frac{19}{96}\) c) \(\frac{1}{6}\)

- Count ordered triples for the small possible sums. - Use the complement of no \(8\). - For all different values, count the remaining choices at each roll.

1. There are \(8^3=512\) equally likely ordered outcomes. Sums less than \(6\) are \(3\), \(4\), and \(5\), occurring in \(1\), \(3\), and \(6\) ordered outcomes, respectively. Thus, \(P(S<6)=\frac{10}{512}=\frac{5}{256}\). 2. The probability of no \(8\) in three rolls is \(\left(\frac{7}{8}\right)^3\). Therefore, \(P(\text{at least one }8)=1-\frac{343}{512}=\frac{169}{512}\). 3. The first result can be any of \(8\) values, the second any of the remaining \(7\), and the third any of the remaining \(6\). Thus, \(P(\text{all different})=\frac{8\cdot7\cdot6}{8^3}=\frac{21}{32}\).

1. \(\frac{5}{256}\approx 0.01953\) 2. \(\frac{169}{512}\approx 0.33008\) 3. \(\frac{21}{32}=0.65625\)

- Use the \(36\) ordered pairs as the equally likely outcomes. - Count pairs with each prime sum. - Compare how many pairs produce different sums.

1. The \(36\) ordered pairs are equally likely. 2. Prime sums are \(2,3,5,7,11\). Counting their ordered pairs gives \(1+2+4+6+2=15\), so \(P(E_1)=\frac{15}{36}=\frac{5}{12}\). 3. Exactly \(15\) ordered pairs have a product divisible by \(4\), so \(P(E_2)=\frac{15}{36}=\frac{5}{12}\). 4. The student’s claim is false. The ordered pairs are equally likely, but the sums are not. For example, a sum of \(2\) has one ordered pair, while a sum of \(7\) has six.

1. \(P(E_1)=\frac{5}{12}\) and \(P(E_2)=\frac{5}{12}\) 2. The claim is false because different sums occur in different numbers of equally likely ordered pairs.

- Count ordered pairs with the required difference. - Use the complement that both results are at most \(6\). - Remember that relative frequencies usually become more stable as the number of simulation trials increases.

1. There are \(64\) equally likely ordered pairs. An absolute difference of \(2\) occurs in \(12\) pairs, so \(P=\frac{12}{64}=\frac{3}{16}=0.1875\). 2. The complement is that both results are from \(1\) through \(6\), which has \(6\cdot6=36\) outcomes. Thus, \(P=1-\frac{36}{64}=\frac{7}{16}=0.4375\). 3. The relative frequency is \(\frac{4120}{10000}=0.412\), which is \(0.0255\) below \(0.4375\). Because \(10{,}000\) is a large number of trials, relative frequencies are generally expected to be fairly stable. This gap is large enough to justify checking the simulation method and recorded count, although the discrepancy alone does not prove that an error occurred.

1. \(\frac{3}{16}=0.1875\) 2. \(\frac{7}{16}=0.4375\) 3. The relative frequency is \(0.412\), which is \(0.0255\) below the theoretical probability. The simulation method and recorded count should be checked, but the discrepancy alone does not prove an error.

- Use the \(144\) ordered pairs as the sample space. - List favorable pairs systematically. - For square products, consider the prime-factor parity of each factor.

1. There are \(12\cdot12=144\) equally likely ordered pairs. 2. A sum of \(15\) occurs in \(10\) ordered pairs, so \(P=\frac{10}{144}=\frac{5}{72}\). 3. Exactly \(22\) ordered pairs have a perfect-square product, so \(P=\frac{22}{144}=\frac{11}{72}\). 4. An absolute difference of at least \(10\) occurs for \((1, 11),(1, 12),(2, 12)\) and their reverses, giving \(6\) outcomes. Thus, \(P=\frac{6}{144}=\frac{1}{24}\). 5. Prime sums from \(2\) through \(24\) are \(2,3,5,7,11,13,17,19,23\). They occur in \(1+2+4+6+10+12+8+6+2=51\) ordered pairs, so \(P=\frac{51}{144}=\frac{17}{48}\).

1. \(\frac{5}{72}\approx 0.06944\) 2. \(\frac{11}{72}\approx 0.15278\) 3. \(\frac{1}{24}\approx 0.04167\) 4. \(\frac{17}{48}\approx 0.35417\)